33 replies to this topic

[bonanova]

It is a beginner-level bit of mathematical magic to prove that 1=2.
 
This demonstration probably occurs multiple times in the history of this forum.
Its fallacy depends on embedding, and cleverly hiding somewhere in the "analysis," a division by zero.
Which of course is not permitted. So regarding this matter our world is still a safe abode.
 
Nonetheless, this equality still has a life for puzzle solvers.
 
By the application of a familiar proof, the addition of a single well-known mathematical symbol,
and without resorting to prohibited mathematical operations, 1=2 can still be shown to be true.
 
Have fun, and please use spoilers. 

[vistaptb]

Spoiler for Here's one

Put a slash through the equal sign?

[phil1882]

Spoiler for

the most familiar proof i can think of is a^2 +b^2 = c^2.

if i take the logarithm of both sides; i get 

log(a^2 +b^2) = log(c^2)

log(a^2 +b^2) = 2*log©

log(a^2 +b^2)/log© = 2

 

how to show this is also equal to 1 however has me stymied.

[superprismatic]

Spoiler for a solid induction proof

We will prove a more general result using a simple induction argument.
We prove: Any set of numbers are all equal.

Proof:
(Base Case) -- If we have a set consisting of one number, it is clear
that all the numbers in this set are equal.
(Inductive Step) -- We assume that any set of N numbers are equal.
Now, we will show that any set of N+1 numbers are all equal. Let us
take one number out of the set of N+1. We are now left with a set
of N numbers and, by our induction hypothesis, all N of these are
equal. So, we put the number we took out (call this number X)
back in the set to again make it a set of N+1 numbers. We now
take a different number out of this set (call this number Y).
Now, we once again have a set of N numbers which, by the induction
hypothesis, have all equal numbers. The only way this can happen
is if X=Y because each of X and Y are equal to the other numbers in
the set. So, any set of N+1 numbers must all be equal.
(Therefore) By induction, any set of numbers must contain numbers
which are all equal.

In particular, the set {17.3,1,-6,2} must contain numbers which are all
equal. Thus 1=2.

[bushindo]

Spoiler for a solid=solid induction=induction proof'=proof

We will prove a more general result using a simple induction argument.
We prove: Any set of numbers are all equal.

Proof:
(Base Case) -- If we have a set consisting of one number, it is clear
that all the numbers in this set are equal.
(Inductive Step) -- We assume that any set of N numbers are equal.
Now, we will show that any set of N+1 numbers are all equal. Let us
take one number out of the set of N+1. We are now left with a set
of N numbers and, by our induction hypothesis, all N of these are
equal. So, we put the number we took out (call this number X)
back in the set to again make it a set of N+1 numbers. We now
take a different number out of this set (call this number Y).
Now, we once again have a set of N numbers which, by the induction
hypothesis, have all equal numbers. The only way this can happen
is if X=Y because each of X and Y are equal to the other numbers in
the set. So, any set of N+1 numbers must all be equal.
(Therefore) By induction, any set of numbers must contain numbers
which are all equal.

In particular, the set {1,2} must contain numbers which are all
equal. Thus 1=2.

Counter example to this proof =)

Spoiler for

We use the same set- {1, 2}. Following the reasoning above, then
X = 1
Y = 2
but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain.

[superprismatic]

 

Spoiler for a solid=solid induction=induction proof'=proof

We will prove a more general result using a simple induction argument.
We prove: Any set of numbers are all equal.

Proof:
(Base Case) -- If we have a set consisting of one number, it is clear
that all the numbers in this set are equal.
(Inductive Step) -- We assume that any set of N numbers are equal.
Now, we will show that any set of N+1 numbers are all equal. Let us
take one number out of the set of N+1. We are now left with a set
of N numbers and, by our induction hypothesis, all N of these are
equal. So, we put the number we took out (call this number X)
back in the set to again make it a set of N+1 numbers. We now
take a different number out of this set (call this number Y).
Now, we once again have a set of N numbers which, by the induction
hypothesis, have all equal numbers. The only way this can happen
is if X=Y because each of X and Y are equal to the other numbers in
the set. So, any set of N+1 numbers must all be equal.
(Therefore) By induction, any set of numbers must contain numbers
which are all equal.

In particular, the set {1,2} must contain numbers which are all
equal. Thus 1=2.

Counter example to this proof =)

Spoiler for

We use the same set- {1, 2}. Following the reasoning above, then
X = 1
Y = 2
but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain.

 

Spoiler for but

it's the ONLY weak link....Give a guy a little break here!

[bonanova]

Good thinking so far, but it's not as complicated as it looks.

You'll just have to go a bit more outside the box.

[bonanova]

Spoiler for Here's one

Put a slash through the equal sign?

Actually not bad.
But not the solution I had in mind.

[superprismatic]

Spoiler for answer the second

Let T be the square root of 2. Consider the infinite continued
exponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearly
have X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4
because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets
1=2. QED

[superprismatic]

Spoiler for answer the third

Start with the trig identity
cos(2x)=cos²(x)-sin²(x)
re-arrange the equation to
cos(2x)+sin²(x)=cos²(x)
take square roots
sqrt(cos(2x)+sin²(x))=cos(x)
divide by sqrt(2) to get
sqrt(cos(2x)+sin²(x))/sqrt(2)=cos(x)/sqrt(2)
add (3/2) to both sides
(3/2)+sqrt(cos(2x)+sin²(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2)
evaluate at x=(3*pi/4) to get
2=1.
Voila!